TITLE: Inverse function theorem in complex analysis
QUESTION [6 upvotes]: I was wondering whether an inverse function theorem in the complex numbers exists?
I mean, in the real numbers we have that if the derivative of a function is non zero, then  the inverse function is also differentiable in a local region. Is there anything that is related to this for complex functions? Or how would one show that the $Ln$ defined as the inverse function of the exponential function is differentiable?

REPLY [4 votes]: Let $f(z):D \to \mathbb{C}$ be holomorphic in a domain $D$ of the complex plane. and suppose that $f'(z_0) \neq 0$, then it is locally invertible. 
Because of the following, let $f(z)=u(x,y)+iv(x,y)$, then we can think of this function as a function from $f(x,y) : \mathbb{R}^2 \to \mathbb{R}^2$ where $f(x,y)=\left(u(x,y),v(x,y)\right)$. now using the inverse function theorem for functions from $\mathbb{R}^n$ to $\mathbb{R}^2$, we form the Jacobian of this function: $$ J_f(x,y) = \det \begin{pmatrix} u_x & u_y \\
v_x & v_y
 \end{pmatrix} = u_xv_y-u_yv_x$$ and note that the condition of nonzero Jacobian at the point $(x_0,y_0) \equiv z_0$ is the same as $f'(z_0) \neq 0$. Because by Cauchy-Riemann equations we have $$ 0 \neq  u_x(x_0,y_0)v_y(x_0,y_0)-u_y(x_0,y_0)v_x(x_0,y_0) = \left( u_x(x_0,y_0) \right)^2+\left( v_x(x_0,y_0) \right)^2= |f'(z_0)|^2 \ \ \blacksquare $$